Harry potter numbers

[Romy Aran]

Hi all, 

First post here. Nice to meet you all! Thank you Neil for telling me about this group. 

I noticed the following very simple pattern, using the 9 3/4 number from Harry Potter as inspiration. Moving the mixed fraction around to 3/4 9 and "naively" bringing the 9 to the numerator, we get 39/4, which is equivalent to 9 3/4. Given an integer n, I wondered what fractions when paired with the integer produces a mixed fraction with this property. 

Here are some that I found like 9 3/4: 

Thought I'd put this out there for your thoughts. 

Best, Romy 

[Dave Consiglio]

Hi Romy,

Welcome! It's an interesting idea. I'm curious - what do you think about mixing the numbers around? For example, for digits a,b,c,d would acb/d = a cb/d be permitted? What about bac/d = a cb/d? 

Dave

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[Daniel Mondot]

For " acb/d = a cb/d " to work you probably need d=100, and then you have 810 solutions (with a,b,c being single digits), so that doesn't seem too interesting to me...

Daniel.

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[Dave Consiglio]

Good point. I guess the question is how to define this sequence such that trivial solutions like that are excluded. 

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[Giorgos Kalogeropoulos]

For the initial case a + b/c here are some results of how many fractions exist with (b and c) < 10^n

a   ->   <10,       <10^2,           <10^3,         <10^4,           <10^5            <10^6

1           0,            10,                  110,           1110,           11110,        111110, 

2           0,             5,                     55,             555,             5555,           55555, 

3           2,            32,                   332,           3332,          33332,        333332,

4           0,             5,                      55,             555,             5555,          55555,

5           0,             8,                      88,             888,             8888,          88888, 

6           1,           16,                    166,           1666,          16666,        166666, 

7           0,             8,                      94,              951,            9523,         95237,  (interesting)

8           0,             5,                      55,              555,            5555,         55555}, 

9           8,           98,                    998,            9998,         99998,         999998

Giorgos

[M F Hasler]

On Thu, May 1, 2025 at 9:42 AM Romy Aran <romy...@gmail.com> wrote:

Hi all, 

First post here. Nice to meet you all! Thank you Neil for telling me about this group. 

I noticed the following very simple pattern, using the 9 3/4 number from Harry Potter as inspiration. Moving the mixed fraction around to 3/4 9 and "naively" bringing the 9 to the numerator, we get 39/4, which is equivalent to 9 3/4. 

Given an integer n, I wondered what fractions when paired with the integer produces a mixed fraction with this property. 

According to this initial message, unless I miss something, a definition could be something like:

"Integer solutions (x,y,z) to

x + y/z = concat( y, x ) / z , with 0 < y < z ; gcd(y, z) = 1."

We find that this amounts to solve

xz + y = y*10^Lx + x, where Lx is the number of digits of x.

<=>  x * (z-1) = y * (10^Lx  - 1)

<=>  x  = 9 y / (z-1) * R(Lx),  where R(n)=(10^n-1)/9 is the n-th repunit.

So we always have the trivial solutions  y = z-1,  x = 9 * R(n), n=1,2,3,...

9 + 1/2 = 19 / 2 ,  99 + 1/2 = 199 / 2 , ...

9 + 2/3 = 29 / 3 , 99 + 2/3 = 299 / 3 , ...

9 + 3/4 = 39 / 4 , 99 + 3/4 = 399 / 4 , ...

(also for multi-digit numbers, say

9...9 + 10/11 = 109...9 / 11 ,

9...9 + 11/12 = 119...9 / 12 ,   

9...9 + 12/13 = 129...9 / 13 ,  etc.)

but, e.g., for z = 4 we also have

3 + 1/4 = 13 / 4 ,  33 + 1/4 = 133 / 4 ,  333 + 1/4 = 1333 / 4 ,  ...

(The solutions  6...6 + 2/4 = 26...6 / 4  don't satisfy  gcd(y,z)=1.)

For z = 7 we also have (in addition to y = 6, x = 9...9) :

3...3 + 2/7 = 23...3 / 7 , and  6...6 + 4/7 = 46...6 / 7.

For z = 8, in addition to y = 7, x = 9...9, we have the more interesting

142857 + 1/8 = 1142857 / 8 ,  142857 142857 + 1/8 = 1142857 142857 / 8, 

142857 142857 142857 + 1/8 = 1142857 142857 142857 / 8, ...

428571 + 3/8 = 3428571 / 8 ,  428571 428571 + 3/8 = 3428571 428571 / 8, 

428571 428571 428571 + 3/8 = 3428571 428571 428571 / 8, ...

714285 + 5/8 = 5714285 / 8 ,  714285 714285 + 5/8 = 5714285 714285 / 8,

714285 714285 714285 + 5/8 = 5714285 714285 714285 / 8, ...

(and similar for even numerators, which don't qualify). 

Obviously, these are the repeated digits in 1/7.

In general, the larger x values will be some periodic part of 1/(z-1), see A036275,

and A060284, which however list only one representative [without "rotation"].

We note that for any solution (x,y,z), concatenation of x with itself gives again a solution, so we should only consider "primitive" solutions, excluding larger x that are repeated concatenations of smaller solutions x.

I have tentatively submitted https://oeis.org/draft/A383188 : 

Irregular table, read by rows, where row z = 2, 3, 4, ... lists pairs (y, x) such that x + y/z = concat(y, x)/z with 0 < y < z, gcd(y, z) = 1, and primitive x, cf. comments.

I chose this format for several reasons, among others that way most (or maybe all?) rows end with y = z-1, x = 9, which makes it easy to read the flattened sequence as a table. Also, (y, x)  <=>  concat(y, x) / z = x + y/z.

It was also more natural from my initial approach, scanning all z > y > 0, z=2,3,4... for possible x.

But I agree that an infinite square table like in the O.P., 

with rows  x = 1, 2, 3..., might be meaningful and natural, too.

- Maximilian

[Romy Aran]

Thank you Dave! This is an interesting direction. I personally haven't explored cases with more than three coefficients but I'm curious what you get if you go further. 

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-Romy Aran-

Harvard College Class of 2025

[Romy Aran]

Wow!! Absolutely amazing! Still working on understanding your generalization to multi-digit values of x. 

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